codechef Dynamic GCD [树链剖分 gcd]

Dynamic GCD

题意:一棵树,字词树链加,树链gcd


根据\(gcd(a,b)=gcd(a,a-b)\)

得到\(gcd(a_1, a_2, ..., a_i) = gcd(a_1, a_1-a_2, a_2-a_3,...)\)

同时维护原序列和差分序列就行了

无脑树剖,分成几段。不需要轻儿子的差分值。

注意最后答案取一下绝对值!!!

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 5e4+5;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}
#define mid ((l+r)>>1)
#define lc x<<1
#define rc x<<1|1
#define lson lc, l, mid
#define rson rc, mid+1, r

int n, Q, a[N], u, v, d;
char s[5];
struct edge {int v, ne;} e[N<<1];
int cnt, h[N];
inline void ins(int u, int v) {
	e[++cnt] = (edge) {v, h[u]}; h[u] = cnt;
	e[++cnt] = (edge) {u, h[v]}; h[v] = cnt;
}
int dfn[N], dfc, size[N], mx[N], fa[N], deep[N], top[N];
void dfs(int u) {
	size[u] = 1;
	for(int i=h[u]; i; i=e[i].ne) {
		int v = e[i].v;
		if(v == fa[u]) continue;
		fa[v] = u; deep[v] = deep[u]+1;
		dfs(v);
		size[u] += size[v];
		if(size[v] > size[mx[u]]) mx[u] = v;
	}
}
void dfs(int u, int anc) {
	dfn[u] = ++dfc; top[u] = anc;
	if(mx[u]) dfs(mx[u], anc);
	for(int i=h[u]; i; i=e[i].ne) {
		int v = e[i].v;
		if(v != fa[u] && v != mx[u]) dfs(v, v);
	}
}

namespace t1 {
	int c[N];
	void add(int p, int v) {for(; p<=n; p+=p&-p) c[p]+=v;}
	int que(int p) {int ans=0; for(; p; p-=p&-p) ans+=c[p]; return ans;}
	void add(int l, int r, int d) {
		add(l, d); add(r+1, -d);
	}
}

int gcd(int a, int b) {return !b ? a : gcd(b, a % b);}
namespace t2 {
	int t[N<<2];
	void build(int x, int l, int r) {
		if(l == r) t[x] = a[l];
		else {
			build(lson); build(rson);
			t[x] = gcd(t[lc], t[rc]);
		}
	}
	void add(int x, int l, int r, int p, int v) {
		if(l == r) t[x] += v;
		else {
			if(p <= mid) add(lson, p, v);
			else add(rson, p, v);
			t[x] = gcd(t[lc], t[rc]);
		}
	}
	int que(int x, int l, int r, int ql, int qr) {
		if(ql > qr) return 0;
		if(ql <= l && r <= qr) return t[x];
		else {
			int ans = 0;
			if(ql <= mid) ans = gcd(ans, que(lson, ql, qr));
			if(mid < qr)  ans = gcd(ans, que(rson, ql, qr));
			return ans;
		}
	}
}

int seg_gcd(int x, int y) {
	int l = dfn[x], r = dfn[y];
	int a = t1::que(l), b = t2::que(1, 1, n, l+1, r);
	return gcd(a, b);
}
void quer(int x, int y) {
	int ans = 0;
	while(top[x] != top[y]) {
		if(deep[top[x]] < deep[top[y]]) swap(x, y);
		ans = gcd(ans, seg_gcd(top[x], x));
		x = fa[top[x]];
	}
	if(dfn[x] > dfn[y]) swap(x, y);
	ans = gcd(ans, seg_gcd(x, y));
	if(ans < 0) ans = -ans;
	printf("%d\n", ans);
}
void seg_add(int x, int y, int d) {
	int l = dfn[x], r = dfn[y];
	t1::add(l, r, d);
	t2::add(1, 1, n, l, -d);
	if(mx[y]) t2::add(1, 1, n, dfn[mx[y]], d);
}
void add(int x, int y, int d) { //printf("add %d %d  %d\n", x, y, d);
	while(top[x] != top[y]) {
		if(deep[top[x]] < deep[top[y]]) swap(x, y);
		seg_add(top[x], x, d);
		x = fa[top[x]];
	}
	if(dfn[x] > dfn[y]) swap(x, y);
	seg_add(x, y, d);
}
int main() {
	//freopen("in", "r", stdin);
	n = read();
	for(int i=1; i<n; i++) ins(read()+1, read()+1);
	dfs(1); dfs(1, 1);
	//for(int i=1; i<=n; i++) printf("look %d  %d   %d %d\n", i, dfn[i], fa[i], mx[i]);
	for(int i=1; i<=n; i++) a[dfn[i]] = read();
	for(int i=1; i<=n; i++) t1::add(i, i, a[i]);// printf("%d ", a[i]); puts("");
	for(int i=n; i>=1; i--) a[i] = a[i-1] - a[i];
	t2::build(1, 1, n);
	Q = read();
	while(Q--) {
		scanf("%s", s); u = read()+1; v = read()+1;
		//printf("--------------------------%s  %d %d\n", s, u, v);
		if(s[0] == 'F') quer(u, v);
		else d = read(), add(u, v, d);
		//for(int i=1; i<=n; i++) printf("%d ", t1::que(i)); puts(" val");
		//printf("seg_gcd %d\n", seg_gcd(1, 2));
	}
}
posted @ 2017-05-05 19:29  Candy?  阅读(428)  评论(0编辑  收藏  举报